Numbers that are divisible by the number of primes smaller than them Let $\pi(n)$ denote the number of primes less than or equal to $n$ (a.k.a the prime-counting function).
For certain values of $n$, the value of $\frac{n}{\pi(n)}$ is integer.
Here are the first few examples:


*

*$n=  8,\pi(n)= 4,\frac{n}{\pi(n)}=2$

*$n= 27,\pi(n)= 9,\frac{n}{\pi(n)}=3$

*$n= 30,\pi(n)=10,\frac{n}{\pi(n)}=3$

*$n= 33,\pi(n)=11,\frac{n}{\pi(n)}=3$

*$n= 96,\pi(n)=24,\frac{n}{\pi(n)}=4$

*$n=100,\pi(n)=25,\frac{n}{\pi(n)}=4$

*$n=120,\pi(n)=30,\frac{n}{\pi(n)}=4$

*$n=330,\pi(n)=66,\frac{n}{\pi(n)}=5$

*$n=335,\pi(n)=67,\frac{n}{\pi(n)}=5$

*$n=340,\pi(n)=68,\frac{n}{\pi(n)}=5$

*$n=350,\pi(n)=70,\frac{n}{\pi(n)}=5$

*$n=355,\pi(n)=71,\frac{n}{\pi(n)}=5$

*$n=360,\pi(n)=72,\frac{n}{\pi(n)}=5$


$\textbf{Has it been proved that }\mathbf{\forall{k>1},\exists{n}:\frac{n}{\pi(n)}=k}$?
Two aspects which "intuitively" support this statement are:


*

*The prime-number theorem, which implies $\frac{n}{\pi(n)}\approx\ln{n}$.

*There seem to be several such values of $n$ for each value of $k$.


But I'm not sure how either one of them can be used in order to establish a proof.
 A: Yes this is proved by S.W.Golomb.
I cannot resist to show a piece of my work:http://arxiv.org/abs/1311.1398
A: With respect to your question if there is, for each $k$, an $n$ such that $\dfrac{n}{\pi(n)}=k$ I think the answer is affirmative. If we consider the function $f(x) = \dfrac{x}{\pi(x)}$, we can prove that $\displaystyle\lim_{x\to +\infty}f(x)=+\infty$ (by the prime number theorem or by more elementary bounds). If $f$ were continuous, as $f(2) = 2$, we should have an intermediate value such that $f(x)=k$. This naive approach doesn't work (as $f$ is not continuous). However, we can do something similar:
We know that there is an $n\in\mathbb{N}$ such that $f(n)<k$ and $f(n+1)\geq k$ (as $f$ tends to infinite, it can't always be bounded by $k$). If $f(n+1)=k$, then we're done. If not, $f(n+1)>k$. This two inequalities can be rewritten as $n < k \pi(n)$ and $n+1 > k\pi(n+1)$. This means that $$k \pi (n+1) < n+1 < k\pi(n) + 1$$  That is $k(\pi(n+1) - \pi(n)) < 1$. This implies that $\pi(n)=\pi(n+1)$. But then, $$k\pi(n+1) = k\pi(n) < n+1 < k\pi(n)+1$$ Absurd, since $n+1$ is an integer in between two consecutive integers.
